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3 years ago

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A factory manager conducted a test on a random sample of 150 products. Of the products tested, 2 were found to be defective. Bas

ed on this information, how many products in a batch of 3000 are likely to be defective?

2 answers:

lions [1.4K]3 years ago

6 0

Answer:

40

Step-by-step explanation:

Write and solve a proportion:

2 / 150 = x / 3000

150x = 6000

x = 40

Phoenix [80]3 years ago

5 0

Answer:

40 products were likely to be defective.

Step-by-step explanation:

A factory worker conducted a test on a random sample of 150 products out of which 2 were found to be defected.

That means percentage of defective products was =

This concludes that 1.33% products were found to be defective.

Now based on this information we will find defective products in a batch of 3000 products.

Number of defective products = 3000×1.33% = (3000).(1.33) over 100

Therefore 40 products were likely to be defective.

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after a party, you have 2/5 of the brownies you made left over. there are 16 left. how many brownies did you make for the party

vodomira [7]

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Answer:

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2 years ago

Add the fraction 1/2+3/18

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3 years ago

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Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m

andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

$A_1 = 67.58 \ in^2$

2

$A_2 =415.4 \ ft^2$

3

$A_3 = 8.48 \ cm^2$

4

$A_4 = 480.38 \ m^2$

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

$A = \frac{\theta}{360} * \pi r^2$

Now at $r_1 = 11 in$ and $\theta_1 = 64^o$

$A_1 = \frac{64}{360} * 3.142 * 11^2$

$A_1 = 67.58 \ in^2$

Now at $r_2 = 20 ft$ in and $\theta_2 = 119 ^o$

$A_2 = \frac{119}{360} * 3.142 * 20^2$

$A_2 =415.4 \ ft^2$

Now at $r_3 = 6.5 cm$ and $\theta_3 = 23 ^o$

$A_3 = \frac{23}{360} * 3.142 * 6.5 ^2$

$A_3 = 8.48 \ cm^2$

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6 0

3 years ago

The Perimeter is x+38

OLga [1]

P=x+38

add all individual sides together (is also the perimeter):
p=(x+3)+(x-5)+x+(x-2)+(x-4)+(x-1)+(x-1)
=7x-10

set both p equal:
x+38=7x-10
38=6x-10
48=6x
8=x

then insert x=8 in all the side/perimeter equations:
p=x+38=8+38=46

sides:
x=8
x-2=8-2=6
x-1=8-1=7
x-4=8-4=4
x-5=8-5=3
x+3=8+3=11

8 0

3 years ago